Answer:
The probability is [tex]P(X >300 ) = 0.97219 [/tex]
Step-by-step explanation:
From the question we are told that
The capacity of an Airliner is k = 300 passengers
The sample size n = 320 passengers
The probability the a randomly selected passenger shows up on to the airport
[tex]p = 0.96[/tex]
Generally the mean is mathematically represented as
[tex]\mu = n* p[/tex]
=> [tex]\mu = 320 * 0.96[/tex]
=> [tex]\mu = 307.2[/tex]
Generally the standard deviation is
[tex]\sigma = \sqrt{n * p * (1 -p ) }[/tex]
=> [tex]\sigma = \sqrt{320 * 0.96 * (1 -0.96 ) }[/tex]
=> [tex]\sigma =3.50 [/tex]
Applying Normal approximation of binomial distribution
Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as
[tex]P(X > k ) = P( \frac{ X -\mu }{\sigma } > \frac{k - \mu}{\sigma } )[/tex]
Here [tex]\frac{ X -\mu }{\sigma } =Z (The \ standardized \ value \ of \ X )[/tex]
=>[tex]P(X >300 ) = P(Z > \frac{300 - 307.2}{3.50} )[/tex]
Now applying continuity correction we have
[tex]P(X >300 ) = P(Z > \frac{[300+0.5] - 307.2}{3.50} )[/tex]
=> [tex]P(X >300 ) = P(Z > \frac{[300.5] - 307.2}{3.50} )[/tex]
=> [tex]P(X >300 ) = P(Z > -1.914 )[/tex]
From the z-table
[tex]P(Z > -1.914 ) = 0.97219[/tex]
So
[tex]P(X >300 ) = 0.97219 [/tex]
